Question

The scalar product of the vector $\hat{i}+\hat{j}+\hat{k}$ with a unit vector along the sum of the vectors $2\hat{i}+4\hat{j}-5\hat{k}$ and $\lambda\hat{i}+2\hat{j}+3\hat{k}$ is equal to 1, then value of $\lambda$ is

• A. 1
• B. 2
• C. 3
• D. 4

Hint:

Logic Tip: In vector equations leading to $\lambda + A = \sqrt{(.......)}$, squaring can sometimes introduce extraneous roots. Always double-check your final answer by substituting $\lambda=1$ back into the scalar product expression: $(3+6-2)/\sqrt{9+40} = 7/\sqrt{49} = 1$. The root is valid.

Answer 1

The Correct Option is A

Concept:
The scalar product (dot product) of vectors $\vec{A}$ and $\vec{B}$ is $A_x B_x + A_y B_y + A_z B_z$. A unit vector $\hat{n}$ along any vector $\vec{V}$ is found by dividing the vector by its magnitude: $\hat{n} = \frac{\vec{V{|\vec{V}|}$.
Step 1: Find the sum of the two given vectors.
Let $\vec{a} = 2\hat{i} + 4\hat{j} - 5\hat{k}$ and $\vec{b} = \lambda\hat{i} + 2\hat{j} + 3\hat{k}$. Their sum, let's call it $\vec{S}$, is: $$\vec{S} = \vec{a} + \vec{b} = (2+\lambda)\hat{i} + (4+2)\hat{j} + (-5+3)\hat{k}$$ $$\vec{S} = (2+\lambda)\hat{i} + 6\hat{j} - 2\hat{k}$$
Step 2: Determine the unit vector along the sum.
The magnitude of the sum vector $\vec{S}$ is: $$|\vec{S}| = \sqrt{(2+\lambda)^2 + 6^2 + (-2)^2} = \sqrt{(2+\lambda)^2 + 36 + 4} = \sqrt{(2+\lambda)^2 + 40}$$ The unit vector $\hat{s}$ along $\vec{S}$ is: $$\hat{s} = \frac{(2+\lambda)\hat{i} + 6\hat{j} - 2\hat{k{\sqrt{(2+\lambda)^2 + 40$$
Step 3: Set up the equation using the given scalar product condition.
We are given that the scalar product of $\vec{v} = \hat{i} + \hat{j} + \hat{k}$ with the unit vector $\hat{s}$ is equal to 1. $$\vec{v} \cdot \hat{s} = 1$$ $$(\hat{i} + \hat{j} + \hat{k}) \cdot \left( \frac{(2+\lambda)\hat{i} + 6\hat{j} - 2\hat{k{\sqrt{(2+\lambda)^2 + 40 \right) = 1$$ Calculate the dot product in the numerator: $$\frac{1(2+\lambda) + 1(6) + 1(-2)}{\sqrt{(2+\lambda)^2 + 40 = 1$$ $$\frac{\lambda + 2 + 6 - 2}{\sqrt{(2+\lambda)^2 + 40 = 1$$ $$\frac{\lambda + 6}{\sqrt{(2+\lambda)^2 + 40 = 1$$
Step 4: Solve for $\lambda$.
Multiply both sides by the denominator: $$\lambda + 6 = \sqrt{(2+\lambda)^2 + 40}$$ Square both sides to remove the radical: $$(\lambda + 6)^2 = (2+\lambda)^2 + 40$$ Expand both sides using $(a+b)^2 = a^2 + 2ab + b^2$: $$\lambda^2 + 12\lambda + 36 = 4 + 4\lambda + \lambda^2 + 40$$ Cancel $\lambda^2$ from both sides and combine constants: $$12\lambda + 36 = 4\lambda + 44$$ $$12\lambda - 4\lambda = 44 - 36$$ $$8\lambda = 8 \implies \lambda = 1$$